Question: Find the zeros of the function. Enter the solutions from least to greatest. $f(x) = (x - 7)^2 - 64$ $\text{lesser }x = $
Explanation: $\begin{aligned} (x - 7)^2 - 64&= 0 \\\\ (x-7)^2&=64 \\\\ \sqrt{(x-7)^2}&=\sqrt{64} \end{aligned}$ $\begin{aligned} x-7&=\pm8 \\\\ x&=\pm8+7 \\ \phantom{(x - 7)^2 - 64}& \\ x=-1&\text{ or }x=15 \end{aligned}$ In conclusion, $\begin{aligned} \text{lesser }x &= -1 \\\\ \text{greater } x &= 15 \end{aligned}$